The number of $$7$$ digit numbers the sum of whose digits is even, is

Question

Infinity Learn · Accepted Answer

A number of the seven digits will be of the form.$$ax1x2x3x4x50$$,$$ax1x2x3x4x51$$,$$ax1x2x3x4x52$$,$$ax1x2x3x4x53$$,…$$ax1x2x3x4x59where$$ a∈$${1$$,$$2$$,$$3$$,…$$9}and$$ $$x1$$,$$x2$$,$$x3$$,$$x4$$,$$x5$$∈$${0$$,$$1$$,$$2$$,$$3$$,…$$9}$$.Since sum of the digits should be even, therefore,if $$a+x1+x2+x3+x4+x5$$ is an even number, then the digit at units place must be $$0$$, $$2$$, $$4$$, $$6$$, $$8$$ and if $$a+x1+x2+x3+x4+x5$$ is an odd number, then the digit at units place will be $$1$$, $$3$$, $$5$$, $$7$$, $$9$$.∴ Required number $$=$$ $$9$$ × $$10$$ × $$10$$ × $$10$$ × $$10$$ × $$10$$ × $$5$$                                $$=$$ $$9$$ × $$105$$ × $$5$$ $$=$$ $$45$$ × $$105$$.

The number of 7 digit numbers the sum of whose digits is even, is

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